JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

3-\frac{x^2}{12} \leq f(x) \leq 3+\frac{x^3}{9}

for all

x \neq 0

, then the value of

\lim \limits_{{h \rightarrow 0}} f(x)

is equal to

View answer and explanation

Correct option: B

Option B (3) is correct because, according to the Sandwich Theorem, the limit of f(x) as x approaches 0 is squeezed between two functions that both approach 3.

More Questions from LIMITS

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\lim \limits_{{x \rightarrow 0}}\left(\frac{1}{x^2}-\operatorname{cosec}^2 x\right)

is equal to

\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{(1-\sin x)^{2 / 3}}=

The value of

\lim\limits_{h \to 0} 2\left\{ {\frac{{\sqrt 3 \sin \left( {\frac{\pi }{6} + h} \right) - \cos \left( {\frac{\pi }{6} + h} \right)}}{{\sqrt 3 h\left( {\sqrt 3 \cos \,h - \sin \,h} \right)}}} \right\}

\lim \limits_{{x \rightarrow \infty}} x\left(e^{1 / x}-1\right)

is equal to If

f(x)=\frac{(x+1)^7 \sqrt{1+x^2}}{\left(x^2-x+1\right)^6}

, then the value of

f^{\prime}(0)

is equal to

\lim\limits_{n \to \infty } {\left( {1 + \sin \frac{a}{n}} \right)^n}

equals